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\(\frac{-2}{1,5}=\frac{-2.2}{1,5.2}=\frac{-4}{3}=\frac{x-1}{3}\)
\(\Rightarrow x-1=-4\)\(\Rightarrow x=-3\)
Vậy \(x=-3\)
Lời giải :
Do \(VT\ge0\forall x;y\)nên ta có hệ :
\(\hept{\begin{cases}\frac{2}{3}-\frac{1}{2}+\frac{3}{4}x=0\\1,5-\frac{11}{17}+\frac{23}{13}y=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{-2}{9}\\y=\frac{-377}{782}\end{cases}}\)
Vậy...
a) \(\frac{2}{3}x+\frac{5}{7}=\frac{3}{10}\)
=> \(\frac{2}{3}x=\frac{3}{10}-\frac{5}{7}\)
=> \(\frac{2}{3}x=-\frac{29}{70}\)
=> \(x=-\frac{29}{70}:\frac{2}{3}\)
=> \(x=-\frac{29}{70}.\frac{3}{2}\)
=> \(x=-\frac{87}{140}\)
b) \(-\frac{21}{13}x+\frac{1}{3}=-\frac{2}{3}\)
=> \(-\frac{21}{13}x=-\frac{2}{3}-\frac{1}{3}\)
=> \(-\frac{21}{13}x=-\frac{3}{3}\)
=> \(-\frac{21}{13}x=1\)
=> \(x=1:\left(-\frac{21}{13}\right)\)
=> \(x=-\frac{13}{21}\)
c) \(\left|x-1,5\right|=2\)
=> \(\left[{}\begin{matrix}x-1,5=2\\x-1,5=-2\end{matrix}\right.=>\left[{}\begin{matrix}x=2+1,5\\x=-2+1,5\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=3,5\\x=-0,5\end{matrix}\right.=>\left[{}\begin{matrix}x=\frac{7}{2}\\x=-\frac{1}{2}\end{matrix}\right.\)(T/M)
d) \(\left|x+\frac{3}{4}\right|-\frac{1}{2}=0\)
=> \(\left|x+\frac{3}{4}\right|=\frac{1}{2}\)
=> \(=>\left[{}\begin{matrix}x+\frac{3}{4}=\frac{1}{2}\\x+\frac{3}{4}=-\frac{1}{2}\end{matrix}\right.=>\left[{}\begin{matrix}x=\frac{1}{2}-\frac{3}{4}\\x=-\frac{1}{2}-\frac{3}{4}\end{matrix}\right.=>\left[{}\begin{matrix}x=-\frac{1}{4}\\x=-\frac{5}{4}\end{matrix}\right.\)(T/M)
HỌC TỐT
a) \(\frac{2}{3}x+\frac{5}{7}=\frac{3}{10}\)
\(\Leftrightarrow\frac{2}{3}x=\frac{3}{10}-\frac{5}{7}\)
\(\Leftrightarrow\frac{2}{3}x=-\frac{29}{70}\)
\(\Leftrightarrow x=-\frac{29}{70}:\frac{2}{3}\)
\(\Leftrightarrow x=-\frac{87}{140}\)
b) \(-\frac{21}{13}x+\frac{1}{3}=-\frac{2}{3}\)
\(\Leftrightarrow-\frac{21}{13}x=-\frac{2}{3}-\frac{1}{3}\)
\(\Leftrightarrow-\frac{21}{13}x=-1\)
\(\Leftrightarrow x=-1:\left(-\frac{21}{13}\right)\)
\(\Leftrightarrow x=\frac{13}{21}\)
c) \(\left|x-1,5\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1,5=2\\x-1,5=-2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=3,5\\x=-0,5\end{matrix}\right.\)
d) \(\left|x+\frac{3}{4}\right|-\frac{1}{2}=0\)
\(\Leftrightarrow\left|x+\frac{3}{4}\right|=\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{3}{4}=\frac{1}{2}\\x+\frac{3}{4}=-\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{1}{4}\\x=\frac{5}{4}\end{matrix}\right.\)
\(1,5.x-1,5=\frac{5}{3}-x\)
\(1,5.x+x=\frac{5}{3}+1,5\)
\(2,5.x=\frac{19}{6}\)
\(x=\frac{19}{6}:2,5\)
\(x=\frac{19}{15}\)
\(a,\frac{3}{4}+\frac{1}{4}:x=\frac{2}{5}\)
\(\Leftrightarrow\frac{1}{4}:x=\frac{2}{5}-\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{4}:x=\frac{8}{20}-\frac{15}{20}\)
\(\Leftrightarrow\frac{1}{4}:x=\frac{-7}{20}\)
\(\Leftrightarrow x=\frac{1}{4}.\frac{20}{-7}\)
\(\Leftrightarrow x=\frac{20}{-28}\)
\(b,|x-2,5|=1,5\)
\(\Leftrightarrow\orbr{\begin{cases}x-2,5=1,5\\x-2,5=-1,5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1,5+2,5\\x=-1,5+2,5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}\)
\(\frac{\left(x+1\right)}{1,5}=\frac{3}{2\left(x+1\right)}\)
\(\Leftrightarrow\left(x+1\right).2\left(x+1\right)=1,5.3\)
\(\Leftrightarrow\left(x+1\right).2x+2=\frac{9}{2}\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=\frac{9}{2}\\2x+2=\frac{9}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{2}\\2x=\frac{5}{2}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{7}{2}\\x=\frac{5}{4}\end{cases}}}\)
\(\frac{x+1}{1,5}=\frac{3}{2\left(x+1\right)}\)
\(\Rightarrow2\left(x+1\right)^2=4,5\)
\(\Rightarrow\left(x+1\right)^2=\frac{9}{4}\)
\(\Rightarrow\orbr{\begin{cases}x+1=\sqrt{\frac{9}{4}}\\x+1=-\sqrt{\frac{9}{4}}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{5}{2}\end{cases}}\)
\(\frac{2x}{3}=\frac{5}{7}\Rightarrow2x.7=3.5\)
\(\frac{2x.7}{14}=\frac{3.5}{14}\)
\(x=\frac{3.5}{14}\)
\(\Rightarrow x=\frac{15}{14}\)